Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+3y &= 1 \\ 3x-2y &= 1\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-2y = -3x+1$ Divide both sides by $-2$ to isolate $y$ $y = {\dfrac{3}{2}x - \dfrac{1}{2}}$ Substitute this expression for $y$ in the first equation. $-8x+3({\dfrac{3}{2}x - \dfrac{1}{2}}) = 1$ $-8x + \dfrac{9}{2}x - \dfrac{3}{2} = 1$ Simplify by combining terms, then solve for $x$ $-\dfrac{7}{2}x - \dfrac{3}{2} = 1$ $-\dfrac{7}{2}x = \dfrac{5}{2}$ $x = -\dfrac{5}{7}$ Substitute $-\dfrac{5}{7}$ for $x$ back into the top equation. $-8( -\dfrac{5}{7})+3y = 1$ $\dfrac{40}{7}+3y = 1$ $3y = -\dfrac{33}{7}$ $y = -\dfrac{11}{7}$ The solution is $\enspace x = -\dfrac{5}{7}, \enspace y = -\dfrac{11}{7}$.